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According to the Artin–Zorn theorem, any finite alternative division ring is a finite field, but I'm interested in the general non-associative (i.e. not necessarily associative) case. Are there any non-associative finite division rings different from finite fields, or are all non-associative finite division rings finite fields?

Edit: We define a not necessarily associative division ring to be a set $S$ equipped with two binary operations $+$ and $\cdot$ such that

  1. $S$ with $+$ is an abelian group
  2. $S/\{0\}$ with $\cdot$ is a loop (unital quasigroup)
  3. $\cdot$ distributes over $+$.

This comes from the following defintion of a division ring: a set $S$ equipped with two binary operations $+$ and $\cdot$ such that

  1. $S$ with $+$ is an abelian group
  2. $S/\{0\}$ with $\cdot$ is a group (associative loop)
  3. $\cdot$ distributes over $+$.
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    What do you even mean by "not-necessarily-associative division ring"? There are several equivalent ways to define inverses in the presence of associativity and without associativity you have to pick one, and the usual one ($a a^{-1} = a^{-1} a = 1$) is pretty useless without associativity! There's no guarantee that $a^{-1}$ is unique and you can't use it to cancel $a$ from a product. So... what's it even good for? – Qiaochu Yuan Aug 09 '20 at 00:17
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    The way I was thinking it as was a set $S$ with two binary operations $+$ and $\cdot$ such that $\cdot$ distributes over $+$ and that $S$ is an abelian group with respect to $+$ and a loop with respect to $\cdot$. Sorry if I wasn't clear enough. –  Aug 11 '20 at 17:02
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    Whoops, yeah I did. –  Aug 11 '20 at 18:13

3 Answers3

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Consider the algebra over $\mathbb{F}_3$ with basis (as a vector space over $\mathbb{F}_3$) the set $\{1,x,x^2\}$ and multiplication given by: \begin{eqnarray*} x(x^2)&=&x+2,\\ (x^2)x&=&1+x+x^2,\\ (x^2)(x^2)&=&x. \end{eqnarray*}

By construction it is finite, has a two-sided identity $1$ and multiplication distributes over addition. The first two equations demonstrate that it is non-associative. Also left or right multiplication by any fixed non-zero element is bijective (see proof below).

I fixed the first equation and did a computer search through the $676$ possibilities for the other two. Of these $14$ came out as having the left and right cancellation property. One of these was of course $\mathbb{F}_{27}$. The other $13$ are non-associative, and of them the above algebra seemed like the nicest.

Proof of left and right cancellation:

It suffices to prove that left multiplication by any non-zero element is injective, as then it must also be surjective and the algebra will not contain non-zero zero-divisors. Thus right multiplication by any non-zero element would also be injective, hence bijective.

Both $x^3-x^2-x-1$ and $x^3-x-2$ are irreducible over $\mathbb{F}_3$ as they have no roots in $\mathbb{F}_3$.

Left multiplication by a non-zero $\mathbb{F}_3$-linear combination $\alpha(x)$ of $1$ and $x$ is the same map as left multiplication by $\alpha(x)$ in $\mathbb{F}_3[x]/(x^3-x-2)\cong\mathbb{F}_{27}$ - hence bijective.

Similarly left multiplication by a non-zero $\mathbb{F}_3$-linear combination $\alpha(y)$ of $1$ and $y=x^2$ is the same map as left multiplication by $\alpha(y)$ in $\mathbb{F}_3[y]/(y^3-y^2-y-1)\cong\mathbb{F}_{27}$ - hence bijective.

Thus without loss of generality, if there is a non-zero left zero-divisor, there will be one of the form $\lambda+x\pm x^2$, for some $\lambda\in \mathbb{F}_3$. Thus it suffices to check that the matrices representing left multiplication by $x\pm x^2$ have no eigenvalues in $\mathbb{F}_3$. The characteristic polynomials of these matrices are:

$$ \left| \begin{array}{ccc} t&2 &1 \\ 2&t+2&1\\ 2&1&t \end{array}\right| = t^3-t^2-t-1 ,\qquad \left| \begin{array}{ccc} t&1 &1 \\ 2&t+1&0\\ 1&0&t \end{array}\right| = t^3+t^2+2 .$$

Neither of these cubics have roots in $\mathbb{F}_3$.

tkf
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We provide a family of examples of finite division algebras, which additionally have a right identity. This does not answer the revised version of the question which asks for a finite division algebra (other than a finite field) which has a two-sided identity.

On any finite field $\mathbb{F}_q$ with $q=p^r$, and $p$ prime, $r>1$, we can define $a\star b= ab^p$. This is non-commutative ($a\star b\neq b\star a \iff a^{-1}b\notin \mathbb{F}_p$), but has the two-sided cancellation property: $$a\star b=0\implies a=0\,\, {\rm or}\,\, b=0,$$ and has a right identity.

tkf
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  • "This does not answer the revised version of the question which asks for a finite division algebra (other than a finite field) which has a two-sided identity." Lol, I forgot rings could be considered as algebras whose vector space basis is just the set ${1}$. –  Aug 12 '20 at 21:57
  • @MadeleineBirchfield By division algebra, I meant over the field $\mathbb{F}_p$ generated additively by your unit. I have added a proof to the other answer, so it now stands independent of computation. Still slightly unsatisfying that it did not occur conceptually. – tkf Aug 12 '20 at 22:14
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Nonassociative division rings with unity are known as semifields. They come up in the coordinatization of projective planes. Their study in the finite case started with

Donald Knuth, Finite semifields and projective planes. J. Algebra 2 (1965), 182-217.

This published version was based on Knuth's 1963 PhD dissertation.

There is a considerable literature on semifields, but note that the term "semifield" is also used in a conflicting sense in other parts of mathematics as an associative semiring with unity in which every nonzero element has a multiplicative inverse. Papers on semifields in the nonassociative sense are generally tagged with the MSC classification 17A35 (Nonassociative division algebras).

Michael Kinyon
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